4.12 grams of Zinc are placed in a crucible with 3.00 grams (excess) of Sulfur. when reaction is complete the product mass is 6.10 grams. What mass of Sulfur should be used in the simplest formula calculation? Find the empirical formula of Zinc Sulfide.

My work...
6.10 g Product - 4.12 g Zn = 1.98 g S

4.12 g Zn • (1 mole Zn)/(65.38 g Zn) = 0.063 moles Zn

1.98 g S • (1 mole S)/(32.06 g S) = 0.062 moles S

1.98 g Sulfur should be used in simplest formula calculation.

Making the simplest empirical formula ZnS because the moles are basically the same so the mole ratio is one to one....

Please let me know if this is right and if not please show me where I went wrong...

the mole ratio is very nearly 1:1, so your answer seems good.

Your work is correct! You have correctly calculated the moles of zinc (Zn) and sulfur (S) based on their respective masses. From there, you determined that the simplest empirical formula of zinc sulfide (ZnS) can be deduced because the mole ratio between Zn and S is approximately 1:1.

Just to reiterate your calculations:

1. Calculate the mass of sulfur (S) by subtracting the mass of zinc (Zn) from the total product mass:
6.10 g (product mass) - 4.12 g (mass of Zn) = 1.98 g (mass of S)

2. Convert the masses of Zn and S to moles by using their respective molar masses:
4.12 g Zn * (1 mole Zn / 65.38 g Zn) = 0.063 moles Zn
1.98 g S * (1 mole S / 32.06 g S) = 0.062 moles S

Since the mole ratios of Zn and S are nearly equal (0.063 moles Zn vs. 0.062 moles S), we can deduce that the simplest empirical formula is ZnS.